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would appreciate some guidance on the following:

Consider the function $z(x,y)=(x+y)\ln(x/y)$.

Show by substitution that $x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=z$.

I have rewritten the equation using basic log rules to try and separate $x$ and $y$, hoping this will shed light on the next step but its still not clear, i.e.

$$z=x\ln(x/y)+y\ln(x/y)$$

$$z=x\ln(x)-x\ln(y)+y\ln(x)-y\ln(y)$$

Thanks in advance.

abnry
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Andy Kellener
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  • If $z=x\ln(x)-x\ln(y)+y\ln(x)-y\ln(y)$, then $\partial z/\partial x=\ln(x/y)+1+y/x$, and $\partial z/\partial y=-x/y+\ln(x/y)-1$. Can you take it from here? –  Feb 08 '14 at 19:50

1 Answers1

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Just differentiate ansd substitute in one side to get the other $$x[\color{red}{\ln x+1}-\ln y+\color{blue}{\frac{y}{x}}]+y[-\frac{x}{y}+\ln x- \ln y-1] \\=x\ln(x)-x\ln(y)+y\ln(x)-y\ln(y)=z$$

Semsem
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  • Thank you both very much. From your responses I have worked out that for ∂z/∂x I should use the product rule on xln(x) and -xln(y) to get ln(x/y)+1+y/x. For the xln(x) I get 1ln(x)+x1/x = ln(x)+1. But, for -xln(y) I get -1ln(y)-x1/y=-ln(y)-x/y. Therefore for ∂z/∂x I get ln(x/y)+1-x/y yet I see in your answers you both have ln(x/y)+1+y/x in ∂z/∂x? What am I missing! Thank you again very much in advance? – Andy Kellener Feb 08 '14 at 20:27
  • I couldn't read your comment, red=derivative of first term, black=2nd, blue=3rd, o=4th. One more thing is that differentiating partially w.r.t $x$, you consider y as constant – Semsem Feb 08 '14 at 20:35
  • @AndyKellener do you accept this or you need it more clear – Semsem Feb 08 '14 at 20:36
  • hi my confusion is over y/x in blue, which I have worked out was derived using the product rule on the −xln(y) part of the equation. I instead get: 1ln(y)-x1/y=-ln(y)-x/y, yet your answer has -ln(y)+y/x? ie you have +x/y while i get -x/y. I can tell a am making a basic error. Thanks. – Andy Kellener Feb 08 '14 at 20:44
  • it is one function since $y$ is considered as a constant – Semsem Feb 08 '14 at 20:45
  • I have got it. Thank you: ∂z/∂x on the −xln(y) part of the equation is just -ln(y), then for the yln(x) part I get y1/x. Then when I bring all the terms in ∂z/∂x together I get ln(x)+1-ln(y)+y1/x which = ln(x/y)+1+y/x. thanks again to both of you.... – Andy Kellener Feb 08 '14 at 21:03
  • I accept...graciously. – Andy Kellener Feb 08 '14 at 21:20